
Re: 2.01  In words and pictures: exposing mythmaticians' misconceptions about epsilondelta theory and problems.
Posted:
Jul 11, 2014 2:04 AM


@dullrich wrote:
> I wasn't saying anything about limits.
Wakeup call: we are talking about limits.
> I was talking about that statement you made. Given that f(x) = x, which by the way you should have explained, the statement "0<x4<delta <=> f(x)4<epsilon" is false.
Wrong.
> Because for example if x = 4 then f(x)4 < epslon is true but 0 < x4 is false.
x CANNOT equal to 4 in that definition! You've been 'teaching' for 30 years. by now you should know this!
> See, _you_ are explaining here why _your_ statements about limits are wrong.
Poppycock.
> Curiously this is exactly the error you made earlier in this same thread in your exchange with YBM.
dullrich, you have sufficient problems dealing with the questions I asked you. So far, you have answered none. Look for thread 1.86 and answer the questions. I am not interested in more dialogue with you. Let's just say that the chances of us having a beer together are virtually null. Don't take this too personally!
> He said something about something you said, and you said no that's not the definition of limit. He didn't say it was.
You did not understand the communication between YBM and me. In fact, neither did he from the looks of things, and if he did, he is just another dishonest reptile, like you.
>Now answer my question:
>> Better example, where the problem is not >> just at the one point: >> >> Let f(x) = sin(x). Then lim_{x>0} f(x) = 0. >> Say epsilon = 1/2. That statement about the >> limit says there exists delta so that >> >> 0 < x < delta => sin(x) < 1/2. >> >> That's true. Now you tell us what delta > 0 makes your version >> 0 < x < delta <=> sin(x) < 1/2 >> true.
That was NOT your question!!!! Sneaky, aren't you! We were talking about infinite limits in that particular series of messages, not finite limits.
But I'll answer your question anyway:
If epsilon = 1/2, then any delta > 1/2 makes my version true.

